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# Chapter 6 - PowerPoint PPT Presentation

Chapter 6. Stability. Stability Definitions. A system is stable if the natural response approaches zero as time approaches infinity A system is unstable if the natural response approaches infinity as time approaches infinity

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Stability

• A system is stable if the natural response approaches zero as time approaches infinity

• A system is unstable if the natural response approaches infinity as time approaches infinity

• A system is marginally stable if the natural response neither decays nor grows but remains constant or oscillates.

• A system is stable if every bounded input yields a bounded output

• A system is unstable if any bounded input yields an unbounded output

Stable systems have closed-loop transfer functions with poles in the left half-plane.Unstable systems have closed-loop transfer functions with at least one pole in the right half-plane, and/or poles of multiplicity greater than one on the imaginary axis Marginally stable systems have closed-loop transfer functions with only imaginary axis poles of multiplicity one and poles in the left half-plane.

Common cause of problems in finding closed-loop poles:

a. original system;b. equivalent system

Using this method we can tell how many closed-loop poles are in the left half-plane, in the right half-plane and on the imaginary axis.

The method requires two steps: (1) Generate the data table (Routh table) and (2) Interpret the table to determine the number of poles in LHP and RHP.

Feedback system for Example 6.1;b. Equivalent closed-loop system

Interpretation of Routh table

The number of roots of the polynomial that are in the right half-plane is equal to the number of sign changes in the first column.

Replace the zero with , the value of is then allowed to approach zero from –ive or +ive side.

ProblemDetermine the stability of the closed-loop transfer function

ProblemDetermine the stability of the closed-loop transfer function

SolutionWrite a polynomial that has the reciprocal roots of the denominator

ProblemDetermine the number of right-half-plane poles in the closed transfer function

Solution: Form an auxiliary polynomial, P(s) using the entries of row above row of zeros as coefficient, then differentiate with respect to s finally use coefficients to replace the rows of zeros and continue the RH procedure.

Root positions to generate even polynomials: A , B, C, or any combination

An entire row of zeros will appear when a purely even or odd polynomial is a factor of the original polynomial.

Even polynomials have roots that are symmetrical about the origin.

If we don’t have a row of zeros, we cannot possibly have roots on the imaginary axis.

ProblemDetermine the number of poles in the right-half-plane, left-half-plan and on the axis for the closed transfer function

ProblemDetermine the number of poles in the right-half-plane, left-half-plan and on the axis for system

Solution: The closed loop transfer function is

2 poles in RHP, 2 poles in LHP no poles on axis The system is unstable

ProblemDetermine the number of poles in the right-half-plane, left-half-plan and on the axis for system

Solution: The closed loop transfer function is

Table for polynomial whose roots are the reciprocal of the original

2 sign changes, 2 poles in RHP system is unstable

ProblemDetermine the number of poles in the right-half-plane, left-half-plan and on the axis for system. Draw conclusions about stability of the closed loop system.

Solution: The closed loop transfer function is

There is a row of zeros in the s5, so the s6 row forms an even polynomial. 2 sign changes from the even polynomial so 2 poles in the RHP and because of symmetry about origin 2 will be in the LHP

The 2 remaining poles are on the axis

ProblemFind the range of gain K for the system that will cuase the system to be stable, unstabel, and marginally stable. Assume K>0.

Solution: The closed loop transfer function is

For K < 1386 the system is stable. For K > 1386 the system is unstable. For K = 1386 we will have entire row of zeros (s row). We form the even polynomial and differentiate and continue, no sign changes from the even polynomial so the 2 roots are on the axis and the system is marginally stable

Problem Factor the polynomial

Solution: from the Routh table we see that the s1 row is a row of zeros. So the even polynomial at the s2 row is since this polynomial is a factor of the original, dividing yields

As the other factor so

Problem Given the system

Find out how many poles in the LHP, RHP and on the axis

Solution: First form (sI-A)

Now find the det (sI-A) =

Routh table for Example 6.11

One sign change, so 1 pole in the LHP and the system is unstable